Question

If $x,y\in (0,30)$ such that
$\left[\frac{x}{3}\right]+\left[\frac{3x}{2}\right]+\left[\frac{y}{2}\right]+\left[\frac{3y}{4}\right]=\frac{11}{6}x+\frac{5}{4}y$
(where $[x]$ denotes greatest integer $\le x$ ), then number of ordered pairs $(x,y)$ is

• A. 0
• B. 2
• C. 4
• D. none of these

Answer 1

Answer: (D)

Let $\{x\}=x-[x]$ denote the fractional part of $x$. Note that $0\le \{x\}<1$. We can write the given equation as
$\frac{x}{3}-\left\{\frac{x}{3}\right\}+\frac{3x}{2}-\left\{\frac{3x}{2}\right\}+\frac{y}{2}-\left\{\frac{y}{2}\right\}+\frac{3y}{4}-\left\{\frac{3y}{4}\right\}=\frac{11}{6}x+\frac{5}{4}y$
$\Rightarrow \left\{\frac{x}{3}\right\}+\left\{\frac{3x}{2}\right\}+\left\{\frac{y}{2}\right\}+\left\{\frac{3y}{4}\right\}=0$
As each number on the L.H.S. lies in the interval $0\le x<1$, we must have
$\left\{\frac{x}{3}\right\}=\left\{\frac{3x}{2}\right\}=\left\{\frac{y}{2}\right\}=\left\{\frac{3y}{4}\right\}=0$
$\Rightarrow \frac{x}{3},\frac{3x}{2},\frac{y}{2}$ and $\frac{3y}{4}$ must be integers.
$\therefore x=6,12,18,24,y=4,8,12,16,20,24,28$
$\Rightarrow$ Number of ordered pairs $(x,y)$ equals $4\times 7=28$.